We are tasked with showing that the sequence $a_n = \left( \frac{(2n)!}{(n!)^2} \right)^{1/n}$ converges to 4 as $n \to \infty$.

The expression $a_n = \left( \frac{(2n)!}{(n!)^2} \right)^{1/n}$ is closely related to the central binomial coefficient, as $\frac{(2n)!}{(n!)^2}$ is the binomial coefficient $\binom{2n}{n}$. Hence, the sequence can be rewritten as:

$a_n = \left( \binom{2n}{n} \right)^{1/n}.$

Step 1: Stirling's Approximation

To find the limit of $a_n$, we use Stirling's approximation for factorials, which is an asymptotic approximation for large $n$. Stirling’s approximation states that:

$n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \quad \text{as} \, n \to \infty.$

Using this approximation for both $(2n)!$ and $(n!)^2$, we have:

$(2n)! \sim \sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}, \quad n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n.$

Step 2: Applying Stirling's Approximation to $a_n$

Now apply Stirling's approximation to the expression $\frac{(2n)!}{(n!)^2}$:

$\frac{(2n)!}{(n!)^2} \sim \frac{\sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}}{\left( \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \right)^2} = \frac{\sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n}}{2 \pi n \left( \frac{n}{e} \right)^{2n}}.$

Simplifying the powers of $n$ and constants:

$\frac{(2n)!}{(n!)^2} \sim \frac{\sqrt{4 \pi n}}{2 \pi n} \cdot 4^n = \frac{2}{\sqrt{\pi n}} \cdot 4^n.$

Step 3: Taking the $n$-th Root

We now take the $n$-th root of the expression:

$a_n = \left( \frac{(2n)!}{(n!)^2} \right)^{1/n} \sim \left( \frac{2}{\sqrt{\pi n}} \cdot 4^n \right)^{1/n} = \left( \frac{2}{\sqrt{\pi n}} \right)^{1/n} \cdot 4.$

As $n \to \infty$, the term $\left( \frac{2}{\sqrt{\pi n}} \right)^{1/n}$ approaches 1, because the $n$-th root of a term that decreases to zero becomes closer and closer to 1. Hence, we have:

$\lim_{n \to \infty} a_n = 4.$

Conclusion

We have shown that:

$\lim_{n \to \infty} a_n = 4.$

Thus, the sequence $a_n = \left( \frac{(2n)!}{(n!)^2} \right)^{1/n}$ converges to 4.

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Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

1. How does Stirling’s approximation help in solving asymptotic problems in combinatorics?
2. Why does $\left( \frac{2}{\sqrt{\pi n}} \right)^{1/n}$ approach 1 as $n \to \infty$?
3. Can we generalize this method to approximate other sequences involving factorials?
4. What are other methods to find limits of sequences without using Stirling’s approximation?
5. How does the central binomial coefficient grow for large $n$?

Tip: Stirling's approximation is a powerful tool for approximating factorials in asymptotic analysis—useful for combinatorics and probability problems!